How do you express # (x^3 +x^2+2x+1) / [(x^2+1) (x^2+2)]# in partial fractions?

Question

1270 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-26T15:29:44

The answer is #=(x)/(x^2+1)+(1)/(x^2+2)#

Let's perform the decomposition into partial fractions

#(x^3+x^2+2x+1)/((x^2+1)(x^2+2))=(Ax+B)/(x^2+1)+(Cx+D)/(x^2+2)#

#=((Ax+B)(x^2+1)+(Cx+D)(x^2+1))/((x^2+1)(x^2+2))#

The denominators are the same, we compare the numerators

#x^3+x^2+2x+1=(Ax+B)(x^2+2)+(Cx+D)(x^2+1)#

Coeficients of #x^3#

#1=A+C#

Let #x=0#, #=>#, #1=2B+D#

Coefficients of #x^2#

#1=B+D#

Coefficients of #x#

#2=2A+C#

Solving for #A#, #B#, #C# and #D# from the 4 equations

#2=2A+1-A#, #=>#, #A=1#

#C=1-A=1-1=0#

#1=2B+1-B#, #=>#, #B=0#

#1=B+D#, #=>#, #D=1#

Therefore,

#(x^3+x^2+2x+1)/((x^2+1)(x^2+2))=(x)/(x^2+1)+(1)/(x^2+2)#